See my post on Discourse:
https://scilab.discourse.group/t/exponentially-modified-gaussian-distribution
S.
On 4/30/24 08:46, Stéphane Mottelet wrote:
Hello,
As I already said, there is no need for using this technique, as an EMG
random variable is the sum of a Gaussian and an exponential. If
Hello,
As I already said, there is no need for using this technique, as an EMG
random variable is the sum of a Gaussian and an exponential. If your CDF
reads as
f = (a/2)*exp((a/2)*((a*σ^2)-2*h)) .* erfc((a*σ^2-h)/σ/sqrt(2))
then the underlying variable is the sum of Gaussian of mean 0,
Very useful. I had done a similar thing with the EMG, but much more clumsy
Heinz
> On 29.04.2024, at 16:41, Federico Miyara wrote:
>
>
> Heinz,
>
> I don't know if this might be useful. The function I'm attaching allows to
> generate random numbers according to any distribution, either
On 29.04.2024, at 09:24, Stéphane Mottelet wrote:
>
> Sorry,
>
> This is just a manifestation of another occurence of the xy problem. You
> told about y (drawing from a given PDF) but the original problem was x :
> drawing a random variable which is the sum of two random variables for
> which we
Sorry,
This is just a manifestation of another occurence of the xy problem. You
told about y (drawing from a given PDF) but the original problem was x :
drawing a random variable which is the sum of two random variables for
which we already have generators. In fact an EMG random variable Z may
Hello,
I think that the best method available for this application (you wil
need a massive number of draws) is the rejection method
(https://en.wikipedia.org/wiki/Rejection_sampling). Let us continue this
discussion on Scilab's Discourse...
S.
On 4/29/24 05:11, Heinz Nabielek wrote:
Colleagues:
bird flight altitude probabilities are given by the exponentially modified
Gaussian distribution EMG f=f(h), in my case
f = (a/2)*exp((a/2)*((a*σ^2)-2*h)) .* erfc((a*σ^2-h)/σ/sqrt(2))
with h=hmeasured-85 in meters, a=1/60m, σ=30m. See:
